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(3x+30)+(x2-40)+(6x)=180
We move all terms to the left:
(3x+30)+(x2-40)+(6x)-(180)=0
We add all the numbers together, and all the variables
(+x^2-40)+(3x+30)+6x-180=0
We add all the numbers together, and all the variables
(+x^2-40)+6x+(3x+30)-180=0
We get rid of parentheses
x^2+6x+3x-40+30-180=0
We add all the numbers together, and all the variables
x^2+9x-190=0
a = 1; b = 9; c = -190;
Δ = b2-4ac
Δ = 92-4·1·(-190)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-29}{2*1}=\frac{-38}{2} =-19 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+29}{2*1}=\frac{20}{2} =10 $
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