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5x^2+21x+6=0
a = 5; b = 21; c = +6;
Δ = b2-4ac
Δ = 212-4·5·6
Δ = 321
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{321}}{2*5}=\frac{-21-\sqrt{321}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{321}}{2*5}=\frac{-21+\sqrt{321}}{10} $
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