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7x^2+21x=3
We move all terms to the left:
7x^2+21x-(3)=0
a = 7; b = 21; c = -3;
Δ = b2-4ac
Δ = 212-4·7·(-3)
Δ = 525
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{525}=\sqrt{25*21}=\sqrt{25}*\sqrt{21}=5\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-5\sqrt{21}}{2*7}=\frac{-21-5\sqrt{21}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+5\sqrt{21}}{2*7}=\frac{-21+5\sqrt{21}}{14} $
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