5t^2+6t-25=0

Solution for 5t^2+6t-25=0 equation:

5t^2+6t-25=0

a = 5; b = 6; c = -25;
Δ = b2-4ac
Δ = 62-4·5·(-25)
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{134}}{2*5}=\frac{-6-2\sqrt{134}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{134}}{2*5}=\frac{-6+2\sqrt{134}}{10}
$