(2a+5)(a-7)=0

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Solution for (2a+5)(a-7)=0 equation:



(2a+5)(a-7)=0
We multiply parentheses ..
(+2a^2-14a+5a-35)=0
We get rid of parentheses
2a^2-14a+5a-35=0
We add all the numbers together, and all the variables
2a^2-9a-35=0
a = 2; b = -9; c = -35;
Δ = b2-4ac
Δ = -92-4·2·(-35)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-19}{2*2}=\frac{-10}{4} =-2+1/2 $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+19}{2*2}=\frac{28}{4} =7 $

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