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7x^2=3x+10
We move all terms to the left:
7x^2-(3x+10)=0
We get rid of parentheses
7x^2-3x-10=0
a = 7; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·7·(-10)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-17}{2*7}=\frac{-14}{14} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+17}{2*7}=\frac{20}{14} =1+3/7 $
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