A portion of the Quadratic Formula proof is shown. Fill in the missing reason. Statements Reasons ax2 + bx + c = 0 Given ax2 + bx = −c Subtract c from both sides of the equation x squared plus b over a times x equals negative c over a Divide both sides of the equation by a x squared plus b over a times x plus the quantity b over 2 times a squared equals negative c over a plus the quantity b over 2 times a squared Complete the square and add the quantity b over 2 times a squared to both sides x squared plus b over a times x plus the quantity b over 2 times a squared equals negative c over a plus b squared over 4 times a squared Square the quantity b over 2 times a on the right side of the equation x squared plus b over a times x plus the quantity b over 2 times a end quantity squared equals negative 4 times a times c over 4 times a squared plus b squared over 4 times a squared Find a common denominator on the right side of the equation x squared plus b over a times x plus the quantity b over 2 times a squared equals b squared minus 4 times a times c all over 4 times a squared Add the fractions together on the right side of the equation quantity x plus b over 2 times a end quantity squared equals b squared minus 4 times a times c all over 4 times a squared ? Rewrite the perfect square trinomial as a binomial squared on the left side of the equation Take the square root of both sides of the equation Multiply both sides of the equation by 2 Square the left side of the equation

Add the fractions together on the right side of the equation.

Step-by-step explanation:

in order to get to the next step, you add the two fractions together that were from the previous step.

Find a common denominator on the right side of the equation

Step-by-step explanation:

I was taking the test and couldn't find the answer to this question anywhere so let me help you out...  

abidemiokin provided a wonderful step-by-step explanation for solving the formula, but left out the actual answer for the question which is what I'm here for. I read through abidemiokin's explanation and reviewed my options before choosing the best answer. As you see in step 5 of my edited version of abidemiokin's explanation the missing step is there. In the provided question we see steps 1,2, 3, and 4 being identical to those in the explanation below. From this, it is clear the next step (step 5) would be the next step, and as it is one of the options we can trust that it is correct.

(If you still have doubts then trust the fact that I took the test and the answer I provided above was indeed correct.)

Given the quadratic equation ax²+bx+c = 0, to derive the quadratic formula from the equation, the following steps must be followed;

ax²+bx+c = 0

Step 1: Subtract c from both sides

ax²+bx+c-c = 0-c

ax²+bx = -c

Step 2: Divide both sides of the equation by a

ax²/a + bx/a = -c/a

x² + bx/a = -c/a

Step 3: Complete the square and add the quantity (b/2a)² times a squared to both sides

x² + bx/a +  (b/2a)² = -c/a +  (b/2a)²

Step 4: Square the quantity  b/2a on the right side of the equation

x² + bx/a +  (b/2a)² =  -c/a +  b²/4a²

Step 5: Find a common denominator on the right side of the equation which is 4a²

x² + bx/a +  (b/2a)² =  -4ac/4a² +  b²/4a²

Step 6: Add the fractions together on the right side of the equation

x² + bx/a +  (b/2a)² =  (-4ac+  b²)/4a²

Note: The fraction at the right-hand side of the equation is to be added together not multiplied as shown in the question.

Step 7: The equation on the left is to be written as a perfect square as shown

(x+b/2a)² =  (-4ac+  b²)/4a²

Step 8: Take the square root of both sides

√(x+b/2a)² = √ (-4ac+  b²)/4a²

(x+b/2a) =  √(-4ac+  b²)/2a

Step 9: subtract b/2a from both sides

x+b/2a - b/2a =  -b/2a + √(-4ac+  b²)/2a

x =  -b/2a + √(-4ac+  b²)/2a

Step 10: Add the fractions together on the right-hand side

x =  -b±√(-4ac+  b²)/2a

This gives the required equation

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )

(-5,-28)

i graphed to on desmos.