-0.035x^2+1.19x+0.04945=0

Solution for -0.035x^2+1.19x+0.04945=0 equation:

-0.035x^2+1.19x+0.04945=0

a = -0.035; b = 1.19; c = +0.04945;
Δ = b2-4ac
Δ = 1.192-4·(-0.035)·0.04945
Δ = 1.423023
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{-(1.19)-\sqrt{1.423023}}{2*-0.035}=\frac{-1.19-\sqrt{1.423023}}{-0.07}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.19)+\sqrt{1.423023}}{2*-0.035}=\frac{-1.19+\sqrt{1.423023}}{-0.07}
$