2h^2+4h=63

Solution for 2h^2+4h=63 equation:

2h^2+4h=63

We move all terms to the left:

2h^2+4h-(63)=0

a = 2; b = 4; c = -63;
Δ = b2-4ac
Δ = 42-4·2·(-63)
Δ = 520
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{130}}{2*2}=\frac{-4-2\sqrt{130}}{4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{130}}{2*2}=\frac{-4+2\sqrt{130}}{4}
$