16x^2+32x+2=10

Solution for 16x^2+32x+2=10 equation:

16x^2+32x+2=10

We move all terms to the left:

16x^2+32x+2-(10)=0

We add all the numbers together, and all the variables

16x^2+32x-8=0

a = 16; b = 32; c = -8;
Δ = b2-4ac
Δ = 322-4·16·(-8)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{6}}{2*16}=\frac{-32-16\sqrt{6}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{6}}{2*16}=\frac{-32+16\sqrt{6}}{32}
$