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2(10x^2-9x-26)=0
We multiply parentheses
20x^2-18x-52=0
a = 20; b = -18; c = -52;
Δ = b2-4ac
Δ = -182-4·20·(-52)
Δ = 4484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4484}=\sqrt{4*1121}=\sqrt{4}*\sqrt{1121}=2\sqrt{1121}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{1121}}{2*20}=\frac{18-2\sqrt{1121}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{1121}}{2*20}=\frac{18+2\sqrt{1121}}{40} $
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