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3x^2=27x-42
We move all terms to the left:
3x^2-(27x-42)=0
We get rid of parentheses
3x^2-27x+42=0
a = 3; b = -27; c = +42;
Δ = b2-4ac
Δ = -272-4·3·42
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-15}{2*3}=\frac{12}{6} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+15}{2*3}=\frac{42}{6} =7 $
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