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x(2x+1)=5
We move all terms to the left:
x(2x+1)-(5)=0
We multiply parentheses
2x^2+x-5=0
a = 2; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·2·(-5)
Δ = 41
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{-(1)-\sqrt{41}}{2*2}=\frac{-1-\sqrt{41}}{4}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*2}=\frac{-1+\sqrt{41}}{4}
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