0.49x^2-13x+40=0

Solution for 0.49x^2-13x+40=0 equation:

0.49x^2-13x+40=0

a = 0.49; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·0.49·40
Δ = 90.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{90.6}}{2*0.49}=\frac{13-\sqrt{90.6}}{0.98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{90.6}}{2*0.49}=\frac{13+\sqrt{90.6}}{0.98}
$