5t^2-49t-10=0

Solution for 5t^2-49t-10=0 equation:

5t^2-49t-10=0

a = 5; b = -49; c = -10;
Δ = b2-4ac
Δ = -492-4·5·(-10)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-51}{2*5}=\frac{-2}{10}
=-1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+51}{2*5}=\frac{100}{10}
=10
$