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7x(2x-3)=21
We move all terms to the left:
7x(2x-3)-(21)=0
We multiply parentheses
14x^2-21x-21=0
a = 14; b = -21; c = -21;
Δ = b2-4ac
Δ = -212-4·14·(-21)
Δ = 1617
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1617}=\sqrt{49*33}=\sqrt{49}*\sqrt{33}=7\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-7\sqrt{33}}{2*14}=\frac{21-7\sqrt{33}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7\sqrt{33}}{2*14}=\frac{21+7\sqrt{33}}{28} $
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