2x^2+10x-19=0

Solution for 2x^2+10x-19=0 equation:

2x^2+10x-19=0

a = 2; b = 10; c = -19;
Δ = b2-4ac
Δ = 102-4·2·(-19)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{7}}{2*2}=\frac{-10-6\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{7}}{2*2}=\frac{-10+6\sqrt{7}}{4}
$