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13x^2-22x+9=0
a = 13; b = -22; c = +9;
Δ = b2-4ac
Δ = -222-4·13·9
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-4}{2*13}=\frac{18}{26} =9/13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+4}{2*13}=\frac{26}{26} =1 $
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