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21x^2-47x+26=0
a = 21; b = -47; c = +26;
Δ = b2-4ac
Δ = -472-4·21·26
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-5}{2*21}=\frac{42}{42} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+5}{2*21}=\frac{52}{42} =1+5/21 $
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