2x^2+16=178

Solution for 2x^2+16=178 equation:

2x^2+16=178

We move all terms to the left:

2x^2+16-(178)=0

We add all the numbers together, and all the variables

2x^2-162=0

a = 2; b = 0; c = -162;
Δ = b2-4ac
Δ = 02-4·2·(-162)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*2}=\frac{-36}{4}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*2}=\frac{36}{4}
=9
$