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3x^2+2x+7=9x+19
We move all terms to the left:
3x^2+2x+7-(9x+19)=0
We get rid of parentheses
3x^2+2x-9x-19+7=0
We add all the numbers together, and all the variables
3x^2-7x-12=0
a = 3; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·3·(-12)
Δ = 193
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{-(-7)-\sqrt{193}}{2*3}=\frac{7-\sqrt{193}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{193}}{2*3}=\frac{7+\sqrt{193}}{6} $
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