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7x^2+21x-4=1
We move all terms to the left:
7x^2+21x-4-(1)=0
We add all the numbers together, and all the variables
7x^2+21x-5=0
a = 7; b = 21; c = -5;
Δ = b2-4ac
Δ = 212-4·7·(-5)
Δ = 581
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{-(21)-\sqrt{581}}{2*7}=\frac{-21-\sqrt{581}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{581}}{2*7}=\frac{-21+\sqrt{581}}{14} $
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