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(21x-5)(10x-3)=147
We move all terms to the left:
(21x-5)(10x-3)-(147)=0
We multiply parentheses ..
(+210x^2-63x-50x+15)-147=0
We get rid of parentheses
210x^2-63x-50x+15-147=0
We add all the numbers together, and all the variables
210x^2-113x-132=0
a = 210; b = -113; c = -132;
Δ = b2-4ac
Δ = -1132-4·210·(-132)
Δ = 123649
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-113)-\sqrt{123649}}{2*210}=\frac{113-\sqrt{123649}}{420} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-113)+\sqrt{123649}}{2*210}=\frac{113+\sqrt{123649}}{420} $
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