2x^2-18=16x

Solution for 2x^2-18=16x equation:

2x^2-18=16x

We move all terms to the left:

2x^2-18-(16x)=0

a = 2; b = -16; c = -18;
Δ = b2-4ac
Δ = -162-4·2·(-18)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*2}=\frac{-4}{4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*2}=\frac{36}{4}
=9
$