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3x(4-x)+3x=6x-97
We move all terms to the left:
3x(4-x)+3x-(6x-97)=0
We add all the numbers together, and all the variables
3x(-1x+4)+3x-(6x-97)=0
We add all the numbers together, and all the variables
3x+3x(-1x+4)-(6x-97)=0
We multiply parentheses
-3x^2+3x+12x-(6x-97)=0
We get rid of parentheses
-3x^2+3x+12x-6x+97=0
We add all the numbers together, and all the variables
-3x^2+9x+97=0
a = -3; b = 9; c = +97;
Δ = b2-4ac
Δ = 92-4·(-3)·97
Δ = 1245
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{-(9)-\sqrt{1245}}{2*-3}=\frac{-9-\sqrt{1245}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{1245}}{2*-3}=\frac{-9+\sqrt{1245}}{-6} $
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