5t^2+2t-55=0

Solution for 5t^2+2t-55=0 equation:

5t^2+2t-55=0

a = 5; b = 2; c = -55;
Δ = b2-4ac
Δ = 22-4·5·(-55)
Δ = 1104
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:

$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{69}}{2*5}=\frac{-2-4\sqrt{69}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{69}}{2*5}=\frac{-2+4\sqrt{69}}{10}
$