5t^2+10t-75=0

Solution for 5t^2+10t-75=0 equation:

5t^2+10t-75=0

a = 5; b = 10; c = -75;
Δ = b2-4ac
Δ = 102-4·5·(-75)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-40}{2*5}=\frac{-50}{10}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+40}{2*5}=\frac{30}{10}
=3
$