t^2+10t+14=-7

Solution for t^2+10t+14=-7 equation:

t^2+10t+14=-7

We move all terms to the left:

t^2+10t+14-(-7)=0

We add all the numbers together, and all the variables

t^2+10t+21=0

a = 1; b = 10; c = +21;
Δ = b2-4ac
Δ = 102-4·1·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*1}=\frac{-14}{2}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*1}=\frac{-6}{2}
=-3
$