7t^2+15t+2=0

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

7t^2+15t+2=0

a = 7; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·7·2
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*7}=\frac{-28}{14}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+13}{2*7}=\frac{-2}{14}
=-1/7
$