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16x^2-20x+6=0
a = 16; b = -20; c = +6;
Δ = b2-4ac
Δ = -202-4·16·6
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*16}=\frac{16}{32} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*16}=\frac{24}{32} =3/4 $
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