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3x^2+6x-18=-5x+7
We move all terms to the left:
3x^2+6x-18-(-5x+7)=0
We get rid of parentheses
3x^2+6x+5x-7-18=0
We add all the numbers together, and all the variables
3x^2+11x-25=0
a = 3; b = 11; c = -25;
Δ = b2-4ac
Δ = 112-4·3·(-25)
Δ = 421
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{-(11)-\sqrt{421}}{2*3}=\frac{-11-\sqrt{421}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{421}}{2*3}=\frac{-11+\sqrt{421}}{6} $
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