Question

solve the equation $2x^2 + 3x - 8 = 4x$ to the nearest tenth. answer attempt 1 out of 2 additional solution no solution $x = $ submit answer

Explanation:

Step1: Rearrange to standard quadratic form

Subtract $4x$ from both sides:
$2x^2 + 3x - 8 - 4x = 0$
Simplify to get:
$2x^2 - x - 8 = 0$

Step2: Identify quadratic coefficients

For $ax^2+bx+c=0$, we have:
$a=2$, $b=-1$, $c=-8$

Step3: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute values:
$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-8)}}{2(2)}$

Step4: Calculate discriminant

Compute the radicand:
$\sqrt{1 + 64}=\sqrt{65}\approx8.062$

Step5: Solve for both x values

First solution (plus sign):
$x=\frac{1+8.062}{4}=\frac{9.062}{4}\approx2.3$
Second solution (minus sign):
$x=\frac{1-8.062}{4}=\frac{-7.062}{4}\approx-1.8$

Answer:

$x\approx2.3$ and $x\approx-1.8$