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(7x+1)(2x-5)=0
We multiply parentheses ..
(+14x^2-35x+2x-5)=0
We get rid of parentheses
14x^2-35x+2x-5=0
We add all the numbers together, and all the variables
14x^2-33x-5=0
a = 14; b = -33; c = -5;
Δ = b2-4ac
Δ = -332-4·14·(-5)
Δ = 1369
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{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-37}{2*14}=\frac{-4}{28} =-1/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+37}{2*14}=\frac{70}{28} =2+1/2 $
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