5t^2+10t-3=0

Solution for 5t^2+10t-3=0 equation:

5t^2+10t-3=0

a = 5; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·5·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*5}=\frac{-10-4\sqrt{10}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*5}=\frac{-10+4\sqrt{10}}{10}
$