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(2x+21)(x-5)=0
We multiply parentheses ..
(+2x^2-10x+21x-105)=0
We get rid of parentheses
2x^2-10x+21x-105=0
We add all the numbers together, and all the variables
2x^2+11x-105=0
a = 2; b = 11; c = -105;
Δ = b2-4ac
Δ = 112-4·2·(-105)
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*2}=\frac{-42}{4} =-10+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*2}=\frac{20}{4} =5 $
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