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2x(3x+3)=10+1-4x+5
We move all terms to the left:
2x(3x+3)-(10+1-4x+5)=0
We add all the numbers together, and all the variables
2x(3x+3)-(-4x+16)=0
We multiply parentheses
6x^2+6x-(-4x+16)=0
We get rid of parentheses
6x^2+6x+4x-16=0
We add all the numbers together, and all the variables
6x^2+10x-16=0
a = 6; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·6·(-16)
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*6}=\frac{-32}{12} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*6}=\frac{12}{12} =1 $
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