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10x^2+26x+16=0
a = 10; b = 26; c = +16;
Δ = b2-4ac
Δ = 262-4·10·16
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-6}{2*10}=\frac{-32}{20} =-1+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+6}{2*10}=\frac{-20}{20} =-1 $
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