2x2+16x+10=0

Solution for 2x2+16x+10=0 equation:

2x^2+16x+10=0

a = 2; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·2·10
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{11}}{2*2}=\frac{-16-4\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{11}}{2*2}=\frac{-16+4\sqrt{11}}{4}
$