t^2+5t-15=0

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



t^2+5t-15=0

a = 1; b = 5; c = -15;
Δ = b2-4ac
Δ = 52-4·1·(-15)
Δ = 85
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{-(5)-\sqrt{85}}{2*1}=\frac{-5-\sqrt{85}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{85}}{2*1}=\frac{-5+\sqrt{85}}{2}
$

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