Please help me and explain how you got your answer!! don’t just give the answer pls

Step-by-step explanation:

2x^2 + 3x =9

when equation like a * (x^2)+ b*x  + c =0

solution like;

x= (-b +sqrt(b^2 - 4ac))/2a

and

x= (-b - sqrt(b^2 - 4ac))/2a

Now consider 2x^2 + 3x -9 =0

so a=2, b=3, c= (-9)

Let first consider  x= (-b + sqrt(b^2 - 4ac))/2a;

x= (-b + sqrt(b^2 - 4ac))/2a

x= ( -3  + sqrt( 3^2 - 4*2*(-9)) )/2 *2

  = 6/4

  = 1.5

Let second consider  x= (-b - sqrt(b^2 - 4ac))/2a;

x= (-b - sqrt(b^2 - 4ac))/2a

x= ( -3  - sqrt( 3^2 - 4*2*(-9)) )/2 *2

  = -12/4

  = -3

$6.42

step-by-step explanation:

step-by-step explanation:

given expression is .

now we need to simplify the given expression .

we know that square root is equivalent to the power 1/2. so that means we can take half of the exponents to get the square root of given expression.

expanding that will give a very very large number so let's keep answer in exponent form.