16x^2+64x-3=0

Solution for 16x^2+64x-3=0 equation:

16x^2+64x-3=0

a = 16; b = 64; c = -3;
Δ = b2-4ac
Δ = 642-4·16·(-3)
Δ = 4288
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:

$\sqrt{\Delta}=\sqrt{4288}=\sqrt{64*67}=\sqrt{64}*\sqrt{67}=8\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{67}}{2*16}=\frac{-64-8\sqrt{67}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{67}}{2*16}=\frac{-64+8\sqrt{67}}{32}
$