t^2+20*t+57=0

Solution for t^2+20*t+57=0 equation:

t^2+20t+57=0

a = 1; b = 20; c = +57;
Δ = b2-4ac
Δ = 202-4·1·57
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{43}}{2*1}=\frac{-20-2\sqrt{43}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{43}}{2*1}=\frac{-20+2\sqrt{43}}{2}
$