4.9t^2-23t+13=0

Solution for 4.9t^2-23t+13=0 equation:

4.9t^2-23t+13=0

a = 4.9; b = -23; c = +13;
Δ = b2-4ac
Δ = -232-4·4.9·13
Δ = 274.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{274.2}}{2*4.9}=\frac{23-\sqrt{274.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{274.2}}{2*4.9}=\frac{23+\sqrt{274.2}}{9.8}
$