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21x^2+5x-35=3x-4x
We move all terms to the left:
21x^2+5x-35-(3x-4x)=0
We add all the numbers together, and all the variables
21x^2+5x-(-1x)-35=0
We get rid of parentheses
21x^2+5x+1x-35=0
We add all the numbers together, and all the variables
21x^2+6x-35=0
a = 21; b = 6; c = -35;
Δ = b2-4ac
Δ = 62-4·21·(-35)
Δ = 2976
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2976}=\sqrt{16*186}=\sqrt{16}*\sqrt{186}=4\sqrt{186}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{186}}{2*21}=\frac{-6-4\sqrt{186}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{186}}{2*21}=\frac{-6+4\sqrt{186}}{42} $
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