2x^2-3x-648=0

Solution for 2x^2-3x-648=0 equation:

2x^2-3x-648=0

a = 2; b = -3; c = -648;
Δ = b2-4ac
Δ = -32-4·2·(-648)
Δ = 5193
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:

$\sqrt{\Delta}=\sqrt{5193}=\sqrt{9*577}=\sqrt{9}*\sqrt{577}=3\sqrt{577}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{577}}{2*2}=\frac{3-3\sqrt{577}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{577}}{2*2}=\frac{3+3\sqrt{577}}{4}
$