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(3x^2)=2x(10+2x)
We move all terms to the left:
(3x^2)-(2x(10+2x))=0
We add all the numbers together, and all the variables
3x^2-(2x(2x+10))=0
We calculate terms in parentheses: -(2x(2x+10)), so:We get rid of parentheses
2x(2x+10)
We multiply parentheses
4x^2+20x
Back to the equation:
-(4x^2+20x)
3x^2-4x^2-20x=0
We add all the numbers together, and all the variables
-1x^2-20x=0
a = -1; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-1)·0
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-1}=\frac{0}{-2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-1}=\frac{40}{-2} =-20 $
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