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(X^2+4X-1)+(5X-3)=0
We get rid of parentheses
X^2+4X+5X-1-3=0
We add all the numbers together, and all the variables
X^2+9X-4=0
a = 1; b = 9; c = -4;
Δ = b2-4ac
Δ = 92-4·1·(-4)
Δ = 97
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{-(9)-\sqrt{97}}{2*1}=\frac{-9-\sqrt{97}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{97}}{2*1}=\frac{-9+\sqrt{97}}{2} $
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