# Solve the inequality -140x+241+19x^2>0

## 19x^2-140x+241>0

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$2.7412>x>4.6272$

## Step by step solution

Problem

$19x^2-140x+241>0$
1

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=19$, $b=-140$ and $c=241$

$x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
2

Substituting the values of the coefficients of the equation in the quadratic formula

$x=\frac{-140\left(-1\right)\pm \sqrt{241\cdot 19\left(-4\right)+{\left(-140\right)}^2}}{19\cdot 2}$
3

Multiply $-1$ times $-140$

$x=\frac{140\pm \sqrt{{\left(-140\right)}^2-18316}}{38}$
4

Calculate the power

$x=\frac{140\pm \sqrt{19600-18316}}{38}$
5

Add the values $19600$ and $-18316$

$x=\frac{140\pm \sqrt{1284}}{38}$
6

Calculate the power

$x=\frac{140\pm 35.8329}{38}$
7

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x_1=\frac{140+ 35.8329}{38}\:\:,\:\:x_2=\frac{140- 35.8329}{38}$
8

Simplifying

$x_1=4.6272,\:x_2=2.7412$
9

Applying the quadratic formula we obtained the two solutions $x_1$ and $x_2$, with which we write the solution interval

$2.7412>x>4.6272$

$2.7412>x>4.6272$

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