16x^2+4x-5=0

Solution for 16x^2+4x-5=0 equation:

16x^2+4x-5=0

a = 16; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·16·(-5)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{21}}{2*16}=\frac{-4-4\sqrt{21}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{21}}{2*16}=\frac{-4+4\sqrt{21}}{32}
$