2x^2+4x-583=0

Solution for 2x^2+4x-583=0 equation:

2x^2+4x-583=0

a = 2; b = 4; c = -583;
Δ = b2-4ac
Δ = 42-4·2·(-583)
Δ = 4680
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:

$\sqrt{\Delta}=\sqrt{4680}=\sqrt{36*130}=\sqrt{36}*\sqrt{130}=6\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{130}}{2*2}=\frac{-4-6\sqrt{130}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{130}}{2*2}=\frac{-4+6\sqrt{130}}{4}
$