4x^2+16x-36=12

Solution for 4x^2+16x-36=12 equation:

4x^2+16x-36=12

We move all terms to the left:

4x^2+16x-36-(12)=0

We add all the numbers together, and all the variables

4x^2+16x-48=0

a = 4; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·4·(-48)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*4}=\frac{-48}{8}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*4}=\frac{16}{8}
=2
$