t^2+30t-50=0

Solution for t^2+30t-50=0 equation:

t^2+30t-50=0

a = 1; b = 30; c = -50;
Δ = b2-4ac
Δ = 302-4·1·(-50)
Δ = 1100
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:

$\sqrt{\Delta}=\sqrt{1100}=\sqrt{100*11}=\sqrt{100}*\sqrt{11}=10\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{11}}{2*1}=\frac{-30-10\sqrt{11}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{11}}{2*1}=\frac{-30+10\sqrt{11}}{2}
$