16t^2+10t-15=0

Solution for 16t^2+10t-15=0 equation:

16t^2+10t-15=0

a = 16; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·16·(-15)
Δ = 1060
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:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{265}}{2*16}=\frac{-10-2\sqrt{265}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{265}}{2*16}=\frac{-10+2\sqrt{265}}{32}
$