2x2+10x=16

Solution for 2x2+10x=16 equation:

2x^2+10x=16

We move all terms to the left:

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

a = 2; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·2·(-16)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{57}}{2*2}=\frac{-10-2\sqrt{57}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{57}}{2*2}=\frac{-10+2\sqrt{57}}{4}
$