2t^2+10t=-5

Solution for 2t^2+10t=-5 equation:

2t^2+10t=-5

We move all terms to the left:

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

We add all the numbers together, and all the variables

2t^2+10t+5=0

a = 2; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·2·5
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{15}}{2*2}=\frac{-10-2\sqrt{15}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{15}}{2*2}=\frac{-10+2\sqrt{15}}{4}
$