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3x^2-11x+30-1x-6=19x-45
We move all terms to the left:
3x^2-11x+30-1x-6-(19x-45)=0
We add all the numbers together, and all the variables
3x^2-12x-(19x-45)+24=0
We get rid of parentheses
3x^2-12x-19x+45+24=0
We add all the numbers together, and all the variables
3x^2-31x+69=0
a = 3; b = -31; c = +69;
Δ = b2-4ac
Δ = -312-4·3·69
Δ = 133
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{-(-31)-\sqrt{133}}{2*3}=\frac{31-\sqrt{133}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{133}}{2*3}=\frac{31+\sqrt{133}}{6} $
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