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0=4x^2-3x^2-6x
We move all terms to the left:
0-(4x^2-3x^2-6x)=0
We add all the numbers together, and all the variables
-(4x^2-3x^2-6x)=0
We get rid of parentheses
-4x^2+3x^2+6x=0
We add all the numbers together, and all the variables
-1x^2+6x=0
a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-1}=\frac{-12}{-2} =+6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2} =0 $
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