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13x^2-28x=42
We move all terms to the left:
13x^2-28x-(42)=0
a = 13; b = -28; c = -42;
Δ = b2-4ac
Δ = -282-4·13·(-42)
Δ = 2968
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2968}=\sqrt{4*742}=\sqrt{4}*\sqrt{742}=2\sqrt{742}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{742}}{2*13}=\frac{28-2\sqrt{742}}{26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{742}}{2*13}=\frac{28+2\sqrt{742}}{26} $
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