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2x^2+3x-4=0x=
We move all terms to the left:
2x^2+3x-4-(0x)=0
We add all the numbers together, and all the variables
2x^2+2x-4=0
a = 2; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·2·(-4)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*2}=\frac{-8}{4} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*2}=\frac{4}{4} =1 $
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