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3x^2-34x+63=0
a = 3; b = -34; c = +63;
Δ = b2-4ac
Δ = -342-4·3·63
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-20}{2*3}=\frac{14}{6} =2+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+20}{2*3}=\frac{54}{6} =9 $
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