How do I solve |5x-3|=-7x?

Solve for x over the real numbers:
abs(5 x-3) = -7 x

Split the equation into two possible cases:
5 x-3 = -7 x or 5 x-3 = 7 x

Add 7 x+3 to both sides:
12 x = 3 or 5 x-3 = 7 x

Divide both sides by 12:
x = 1/4 or 5 x-3 = 7 x

Subtract 7 x-3 from both sides:
x = 1/4 or -2 x = 3

Divide both sides by -2:
x = 1/4 or x = -3/2

abs(5 x-3) ⇒ abs(-3+(5 (-3))/2) = 21/2
-7 x ⇒ -(7 (-3))/2 = 21/2:
So this solution is correct

abs(5 x-3) ⇒ abs(5/4-3) = 7/4
-7 x ⇒ -7/4 = -7/4:
So this solution is incorrect

The solution is:
Answer: |x = -3/2

|5x-3|=-7x,  

$x < 0\ !$

$\begin{array}{|rcll|} \hline |5x-3| &=& -7x \qquad & | \qquad \text{square both sides} \\ (5x-3)^2 &=& (-7x)^2 \\ 25x^2-30x +9 &=& 49x^2 \qquad & | \qquad -(25x^2-30x +9) \\ 0 &=& 49x^2 -25x^2+30x -9 \\ 0 &=& 24x^2 +30x -9 \\ 24x^2 +30x -9 &=& 0 \\\\ x_{1,2} &=& \frac{-30\pm \sqrt{30^2-4\cdot 24\cdot (-9) } }{ 2\cdot 24 } \\ x_{1,2} &=& \frac{-30\pm \sqrt{900+864 } }{ 48 } \\ x_{1,2} &=& \frac{-30\pm \sqrt{ 1764} }{ 48 } \\ x_{1,2} &=& \frac{-30\pm 42 }{ 48 } \\\\ x_1 &=& \frac{-30 + 42 } { 48 } \\ \mathbf{ x_1} &\mathbf{ =}& \mathbf{ 0.25 } \qquad \text{no solution } x > 0 \\\\ x_2 &=& \frac{-30 - 42 } { 48 } \\ \mathbf{ x_2} &\mathbf{ =}& \mathbf{ -1.5 } \\ \hline \end{array}$

The only solution is x = -1.5