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6x^2+3x-4x^2=0
We add all the numbers together, and all the variables
2x^2+3x=0
a = 2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·2·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*2}=\frac{-6}{4} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*2}=\frac{0}{4} =0 $
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