5t^2+16t-12=0

Solution for 5t^2+16t-12=0 equation:

5t^2+16t-12=0

a = 5; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·5·(-12)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{31}}{2*5}=\frac{-16-4\sqrt{31}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{31}}{2*5}=\frac{-16+4\sqrt{31}}{10}
$