5t2+90t-400=0

Solution for 5t2+90t-400=0 equation:

5t^2+90t-400=0

a = 5; b = 90; c = -400;
Δ = b2-4ac
Δ = 902-4·5·(-400)
Δ = 16100
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:

$\sqrt{\Delta}=\sqrt{16100}=\sqrt{100*161}=\sqrt{100}*\sqrt{161}=10\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-10\sqrt{161}}{2*5}=\frac{-90-10\sqrt{161}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+10\sqrt{161}}{2*5}=\frac{-90+10\sqrt{161}}{10}
$