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5x^2+21x=4
We move all terms to the left:
5x^2+21x-(4)=0
a = 5; b = 21; c = -4;
Δ = b2-4ac
Δ = 212-4·5·(-4)
Δ = 521
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{521}}{2*5}=\frac{-21-\sqrt{521}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{521}}{2*5}=\frac{-21+\sqrt{521}}{10} $
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