5t^2+2t-10=0

Solution for 5t^2+2t-10=0 equation:

5t^2+2t-10=0

a = 5; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·5·(-10)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{51}}{2*5}=\frac{-2-2\sqrt{51}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{51}}{2*5}=\frac{-2+2\sqrt{51}}{10}
$