t^2+30t-350=0

Solution for t^2+30t-350=0 equation:

t^2+30t-350=0

a = 1; b = 30; c = -350;
Δ = b2-4ac
Δ = 302-4·1·(-350)
Δ = 2300
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:

$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{23}}{2*1}=\frac{-30-10\sqrt{23}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{23}}{2*1}=\frac{-30+10\sqrt{23}}{2}
$