3t^2+52t-256=0

Solution for 3t^2+52t-256=0 equation:

3t^2+52t-256=0

a = 3; b = 52; c = -256;
Δ = b2-4ac
Δ = 522-4·3·(-256)
Δ = 5776
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5776}=76$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-76}{2*3}=\frac{-128}{6}
=-21+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+76}{2*3}=\frac{24}{6}
=4
$