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4x^2-3x+6=-7x^2+3x+6
We move all terms to the left:
4x^2-3x+6-(-7x^2+3x+6)=0
We get rid of parentheses
4x^2+7x^2-3x-3x-6+6=0
We add all the numbers together, and all the variables
11x^2-6x=0
a = 11; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·11·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*11}=\frac{0}{22} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*11}=\frac{12}{22} =6/11 $
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