Question

solve the equation $6x^2 + 1 = 10x$ to the nearest tenth.
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Explanation:

Step1: Rearrange to standard quadratic form

$6x^2 - 10x + 1 = 0$

Step2: Identify coefficients for quadratic formula

For $ax^2+bx+c=0$, $a=6$, $b=-10$, $c=1$

Step3: Apply quadratic formula

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$x = \frac{10 \pm \sqrt{(-10)^2-4(6)(1)}}{2(6)}$

Step4: Calculate discriminant

$\sqrt{100 - 24} = \sqrt{76} \approx 8.7178$

Step5: Compute two solutions

$x_1 = \frac{10 + 8.7178}{12} \approx \frac{18.7178}{12} \approx 1.6$
$x_2 = \frac{10 - 8.7178}{12} \approx \frac{1.2822}{12} \approx 0.1$

Answer:

$x \approx 1.6$ and $x \approx 0.1$