4t^2+32t-5=0

Solution for 4t^2+32t-5=0 equation:

4t^2+32t-5=0

a = 4; b = 32; c = -5;
Δ = b2-4ac
Δ = 322-4·4·(-5)
Δ = 1104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{69}}{2*4}=\frac{-32-4\sqrt{69}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{69}}{2*4}=\frac{-32+4\sqrt{69}}{8}
$