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6x^2+4x-21=0
a = 6; b = 4; c = -21;
Δ = b2-4ac
Δ = 42-4·6·(-21)
Δ = 520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{130}}{2*6}=\frac{-4-2\sqrt{130}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{130}}{2*6}=\frac{-4+2\sqrt{130}}{12} $
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