5t^2+15t+7=0

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Solution for 5t^2+15t+7=0 equation: 



5t^2+15t+7=0

a = 5; b = 15; c = +7;
Δ = b2-4ac
Δ = 152-4·5·7
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{-(15)-\sqrt{85}}{2*5}=\frac{-15-\sqrt{85}}{10}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{85}}{2*5}=\frac{-15+\sqrt{85}}{10}
$

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