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4c^2+13c-42=0
a = 4; b = 13; c = -42;
Δ = b2-4ac
Δ = 132-4·4·(-42)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{841}=29$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-29}{2*4}=\frac{-42}{8} =-5+1/4 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+29}{2*4}=\frac{16}{8} =2 $
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