-28(x)+6(x)+8(x)^2=112

Solution for -28(x)+6(x)+8(x)^2=112 equation:

-28(x)+6(x)+8(x)^2=112

We move all terms to the left:

-28(x)+6(x)+8(x)^2-(112)=0

determiningTheFunctionDomain
8x^2-28x+6x-112=0

We add all the numbers together, and all the variables

8x^2-22x-112=0

a = 8; b = -22; c = -112;
Δ = b2-4ac
Δ = -222-4·8·(-112)
Δ = 4068
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4068}=\sqrt{36*113}=\sqrt{36}*\sqrt{113}=6\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6\sqrt{113}}{2*8}=\frac{22-6\sqrt{113}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6\sqrt{113}}{2*8}=\frac{22+6\sqrt{113}}{16}
$