t^2+22t=9

Solution for t^2+22t=9 equation:

t^2+22t=9

We move all terms to the left:

t^2+22t-(9)=0

a = 1; b = 22; c = -9;
Δ = b2-4ac
Δ = 222-4·1·(-9)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{130}}{2*1}=\frac{-22-2\sqrt{130}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{130}}{2*1}=\frac{-22+2\sqrt{130}}{2}
$