-5t^2+565t=9

Solution for -5t^2+565t=9 equation:

-5t^2+565t=9

We move all terms to the left:

-5t^2+565t-(9)=0

a = -5; b = 565; c = -9;
Δ = b2-4ac
Δ = 5652-4·(-5)·(-9)
Δ = 319045
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(565)-\sqrt{319045}}{2*-5}=\frac{-565-\sqrt{319045}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(565)+\sqrt{319045}}{2*-5}=\frac{-565+\sqrt{319045}}{-10}
$