The problem is to solve the quadratic equation $10x^2 = -19x + 15$ for $x$.

1. Problem Description

The problem is to solve the quadratic equation $10x^2 = -19x + 15$ for $x$.

2. Solution Steps

First, we rewrite the equation in the standard quadratic form $ax^2 + bx + c = 0$:

$10x^2 = -19x + 15$

$10x^2 + 19x - 15 = 0$

Now, we can solve for $x$ using the quadratic formula:

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

where $a = 10$, $b = 19$, and $c = -15$.

Plugging in the values, we have:

$x = \frac{-19 \pm \sqrt{19^2 - 4(10)(-15)}}{2(10)}$

$x = \frac{-19 \pm \sqrt{361 + 600}}{20}$

$x = \frac{-19 \pm \sqrt{961}}{20}$

$x = \frac{-19 \pm 31}{20}$

So we have two possible solutions:

$x_1 = \frac{-19 + 31}{20} = \frac{12}{20} = \frac{3}{5}$

$x_2 = \frac{-19 - 31}{20} = \frac{-50}{20} = -\frac{5}{2}$

3. Final Answer

The solutions are $x = \frac{3}{5}$ and $x = -\frac{5}{2}$.