4x(2x+5)=7x+19

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Solution for 4x(2x+5)=7x+19 equation:



4x(2x+5)=7x+19
We move all terms to the left:
4x(2x+5)-(7x+19)=0
We multiply parentheses
8x^2+20x-(7x+19)=0
We get rid of parentheses
8x^2+20x-7x-19=0
We add all the numbers together, and all the variables
8x^2+13x-19=0
a = 8; b = 13; c = -19;
Δ = b2-4ac
Δ = 132-4·8·(-19)
Δ = 777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{777}}{2*8}=\frac{-13-\sqrt{777}}{16} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{777}}{2*8}=\frac{-13+\sqrt{777}}{16} $

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