7x^2+32x-19=0

Solution for 7x^2+32x-19=0 equation:

7x^2+32x-19=0

a = 7; b = 32; c = -19;
Δ = b2-4ac
Δ = 322-4·7·(-19)
Δ = 1556
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:

$\sqrt{\Delta}=\sqrt{1556}=\sqrt{4*389}=\sqrt{4}*\sqrt{389}=2\sqrt{389}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{389}}{2*7}=\frac{-32-2\sqrt{389}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{389}}{2*7}=\frac{-32+2\sqrt{389}}{14}
$