(7x-9)(4x+33)=90

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Solution for (7x-9)(4x+33)=90 equation:



(7x-9)(4x+33)=90
We move all terms to the left:
(7x-9)(4x+33)-(90)=0
We multiply parentheses ..
(+28x^2+231x-36x-297)-90=0
We get rid of parentheses
28x^2+231x-36x-297-90=0
We add all the numbers together, and all the variables
28x^2+195x-387=0
a = 28; b = 195; c = -387;
Δ = b2-4ac
Δ = 1952-4·28·(-387)
Δ = 81369
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:
$\sqrt{\Delta}=\sqrt{81369}=\sqrt{9*9041}=\sqrt{9}*\sqrt{9041}=3\sqrt{9041}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(195)-3\sqrt{9041}}{2*28}=\frac{-195-3\sqrt{9041}}{56} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(195)+3\sqrt{9041}}{2*28}=\frac{-195+3\sqrt{9041}}{56} $

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