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2x^2=22x-60
We move all terms to the left:
2x^2-(22x-60)=0
We get rid of parentheses
2x^2-22x+60=0
a = 2; b = -22; c = +60;
Δ = b2-4ac
Δ = -222-4·2·60
Δ = 4
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2}{2*2}=\frac{20}{4} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2}{2*2}=\frac{24}{4} =6 $
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