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9x^2+23x-12=16-6x
We move all terms to the left:
9x^2+23x-12-(16-6x)=0
We add all the numbers together, and all the variables
9x^2+23x-(-6x+16)-12=0
We get rid of parentheses
9x^2+23x+6x-16-12=0
We add all the numbers together, and all the variables
9x^2+29x-28=0
a = 9; b = 29; c = -28;
Δ = b2-4ac
Δ = 292-4·9·(-28)
Δ = 1849
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{1849}=43$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-43}{2*9}=\frac{-72}{18} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+43}{2*9}=\frac{14}{18} =7/9 $
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