16x^2+32x-23=0

Solution for 16x^2+32x-23=0 equation:

16x^2+32x-23=0

a = 16; b = 32; c = -23;
Δ = b2-4ac
Δ = 322-4·16·(-23)
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{39}}{2*16}=\frac{-32-8\sqrt{39}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{39}}{2*16}=\frac{-32+8\sqrt{39}}{32}
$