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22x^2-40x+14=0
a = 22; b = -40; c = +14;
Δ = b2-4ac
Δ = -402-4·22·14
Δ = 368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{23}}{2*22}=\frac{40-4\sqrt{23}}{44} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{23}}{2*22}=\frac{40+4\sqrt{23}}{44} $
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