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(4x^2+2x-3)-(3x^2-3x+4)=0
We get rid of parentheses
4x^2-3x^2+2x+3x-3-4=0
We add all the numbers together, and all the variables
x^2+5x-7=0
a = 1; b = 5; c = -7;
Δ = b2-4ac
Δ = 52-4·1·(-7)
Δ = 53
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{53}}{2*1}=\frac{-5-\sqrt{53}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{53}}{2*1}=\frac{-5+\sqrt{53}}{2} $
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