t^2+10t=-3

Solution for t^2+10t=-3 equation:

t^2+10t=-3

We move all terms to the left:

t^2+10t-(-3)=0

We add all the numbers together, and all the variables

t^2+10t+3=0

a = 1; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·1·3
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{22}}{2*1}=\frac{-10-2\sqrt{22}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{22}}{2*1}=\frac{-10+2\sqrt{22}}{2}
$