5t2+15t=10

Solution for 5t2+15t=10 equation:

5t^2+15t=10

We move all terms to the left:

5t^2+15t-(10)=0

a = 5; b = 15; c = -10;
Δ = b2-4ac
Δ = 152-4·5·(-10)
Δ = 425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{17}}{2*5}=\frac{-15-5\sqrt{17}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{17}}{2*5}=\frac{-15+5\sqrt{17}}{10}
$