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16x^2+26x-12=0
a = 16; b = 26; c = -12;
Δ = b2-4ac
Δ = 262-4·16·(-12)
Δ = 1444
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{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-38}{2*16}=\frac{-64}{32} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+38}{2*16}=\frac{12}{32} =3/8 $
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