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5x^2+26x+21=0
a = 5; b = 26; c = +21;
Δ = b2-4ac
Δ = 262-4·5·21
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-16}{2*5}=\frac{-42}{10} =-4+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+16}{2*5}=\frac{-10}{10} =-1 $
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