5t^2+8t-6=0

Solution for 5t^2+8t-6=0 equation:

5t^2+8t-6=0

a = 5; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·5·(-6)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{46}}{2*5}=\frac{-8-2\sqrt{46}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{46}}{2*5}=\frac{-8+2\sqrt{46}}{10}
$