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2x^2-28x=32
We move all terms to the left:
2x^2-28x-(32)=0
a = 2; b = -28; c = -32;
Δ = b2-4ac
Δ = -282-4·2·(-32)
Δ = 1040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{65}}{2*2}=\frac{28-4\sqrt{65}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{65}}{2*2}=\frac{28+4\sqrt{65}}{4} $
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