-5t^2-15t+12=0

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Solution for -5t^2-15t+12=0 equation: 



-5t^2-15t+12=0

a = -5; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·(-5)·12
Δ = 465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{465}}{2*-5}=\frac{15-\sqrt{465}}{-10}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{465}}{2*-5}=\frac{15+\sqrt{465}}{-10}
$

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