16x^2+12x-20=0

Solution for 16x^2+12x-20=0 equation:

16x^2+12x-20=0

a = 16; b = 12; c = -20;
Δ = b2-4ac
Δ = 122-4·16·(-20)
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{89}}{2*16}=\frac{-12-4\sqrt{89}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{89}}{2*16}=\frac{-12+4\sqrt{89}}{32}
$