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3a^2-26a-40=0
a = 3; b = -26; c = -40;
Δ = b2-4ac
Δ = -262-4·3·(-40)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1156}=34$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-34}{2*3}=\frac{-8}{6} =-1+1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+34}{2*3}=\frac{60}{6} =10 $
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