453-4.72x+0.0114x^2=0

Solution for 453-4.72x+0.0114x^2=0 equation:

453-4.72x+0.0114x^2=0

a = 0.0114; b = -4.72; c = +453;
Δ = b2-4ac
Δ = -4.722-4·0.0114·453
Δ = 1.6216
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{-(-4.72)-\sqrt{1.6216}}{2*0.0114}=\frac{4.72-\sqrt{1.6216}}{0.0228}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.72)+\sqrt{1.6216}}{2*0.0114}=\frac{4.72+\sqrt{1.6216}}{0.0228}
$