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3x^2+21x+10=0
a = 3; b = 21; c = +10;
Δ = b2-4ac
Δ = 212-4·3·10
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*3}=\frac{-21-\sqrt{321}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{321}}{2*3}=\frac{-21+\sqrt{321}}{6} $
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