t^2+9t=10

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

t^2+9t=10

We move all terms to the left:

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

a = 1; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·1·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*1}=\frac{-20}{2}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*1}=\frac{2}{2}
=1
$