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42x^2+98x+56=0
a = 42; b = 98; c = +56;
Δ = b2-4ac
Δ = 982-4·42·56
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-14}{2*42}=\frac{-112}{84} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+14}{2*42}=\frac{-84}{84} =-1 $
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