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12x^2+41x+35=0
a = 12; b = 41; c = +35;
Δ = b2-4ac
Δ = 412-4·12·35
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-1}{2*12}=\frac{-42}{24} =-1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+1}{2*12}=\frac{-40}{24} =-1+2/3 $
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