(16x^2)+68x=-72

Solution for (16x^2)+68x=-72 equation:

(16x^2)+68x=-72

We move all terms to the left:

(16x^2)+68x-(-72)=0

We add all the numbers together, and all the variables

16x^2+68x+72=0

a = 16; b = 68; c = +72;
Δ = b2-4ac
Δ = 682-4·16·72
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4}{2*16}=\frac{-72}{32}
=-2+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4}{2*16}=\frac{-64}{32}
=-2
$