78=(2x+5)(3x-4)

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Solution for 78=(2x+5)(3x-4) equation:



78=(2x+5)(3x-4)
We move all terms to the left:
78-((2x+5)(3x-4))=0
We multiply parentheses ..
-((+6x^2-8x+15x-20))+78=0
We calculate terms in parentheses: -((+6x^2-8x+15x-20)), so:
(+6x^2-8x+15x-20)
We get rid of parentheses
6x^2-8x+15x-20
We add all the numbers together, and all the variables
6x^2+7x-20
Back to the equation:
-(6x^2+7x-20)
We get rid of parentheses
-6x^2-7x+20+78=0
We add all the numbers together, and all the variables
-6x^2-7x+98=0
a = -6; b = -7; c = +98;
Δ = b2-4ac
Δ = -72-4·(-6)·98
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-49}{2*-6}=\frac{-42}{-12} =3+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+49}{2*-6}=\frac{56}{-12} =-4+2/3 $

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