2t2+15t+7=0

Solution for 2t2+15t+7=0 equation:

2t^2+15t+7=0

a = 2; b = 15; c = +7;
Δ = b2-4ac
Δ = 152-4·2·7
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-13}{2*2}=\frac{-28}{4}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+13}{2*2}=\frac{-2}{4}
=-1/2
$