2w^2+9w-63=0

Solution for 2w^2+9w-63=0 equation:

2w^2+9w-63=0

a = 2; b = 9; c = -63;
Δ = b2-4ac
Δ = 92-4·2·(-63)
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{65}}{2*2}=\frac{-9-3\sqrt{65}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{65}}{2*2}=\frac{-9+3\sqrt{65}}{4}
$