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7x(2x-5)=63
We move all terms to the left:
7x(2x-5)-(63)=0
We multiply parentheses
14x^2-35x-63=0
a = 14; b = -35; c = -63;
Δ = b2-4ac
Δ = -352-4·14·(-63)
Δ = 4753
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4753}=\sqrt{49*97}=\sqrt{49}*\sqrt{97}=7\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-7\sqrt{97}}{2*14}=\frac{35-7\sqrt{97}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+7\sqrt{97}}{2*14}=\frac{35+7\sqrt{97}}{28} $
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