-5t^2+15t+90=0

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Solution for -5t^2+15t+90=0 equation:



-5t^2+15t+90=0
a = -5; b = 15; c = +90;
Δ = b2-4ac
Δ = 152-4·(-5)·90
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-45}{2*-5}=\frac{-60}{-10} =+6 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+45}{2*-5}=\frac{30}{-10} =-3 $

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