t^2+10t=9=0

Solution for t^2+10t=9=0 equation:

t^2+10t=9=0

We move all terms to the left:

t^2+10t-(9)=0

a = 1; b = 10; c = -9;
Δ = b2-4ac
Δ = 102-4·1·(-9)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{34}}{2*1}=\frac{-10-2\sqrt{34}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{34}}{2*1}=\frac{-10+2\sqrt{34}}{2}
$