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+7=0

a = 1; b = 10; c = +7;
Δ = b2-4ac
Δ = 102-4·1·7
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{2}}{2*1}=\frac{-10-6\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{2}}{2*1}=\frac{-10+6\sqrt{2}}{2}
$