5t^2+40t=23

Solution for 5t^2+40t=23 equation:

5t^2+40t=23

We move all terms to the left:

5t^2+40t-(23)=0

a = 5; b = 40; c = -23;
Δ = b2-4ac
Δ = 402-4·5·(-23)
Δ = 2060
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:

$\sqrt{\Delta}=\sqrt{2060}=\sqrt{4*515}=\sqrt{4}*\sqrt{515}=2\sqrt{515}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{515}}{2*5}=\frac{-40-2\sqrt{515}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{515}}{2*5}=\frac{-40+2\sqrt{515}}{10}
$