(13-10x)²+15(10x²+7x-26)²=0

To solve the equation \((13-10x)² + 15(10x²+7x-26)² = 0\), we need to understand that a sum of squares is equal to zero only if each square is individually equal to zero.

So, we can set up the following equations:

1. \( (13 - 10x)^2 = 0 \)
2. \( 15(10x^2 + 7x - 26)^2 = 0 \)

Starting with the first equation:

1. \( (13 - 10x)^2 = 0 \) \[ 13 - 10x = 0 \] \[ 10x = 13 \] \[ x = \frac{13}{10} = 1.3 \]

Now for the second equation:

2. \( 15(10x^2 + 7x - 26)^2 = 0 \) Since \(15\) is a non-zero constant, we can simplify this to: \[ (10x^2 + 7x - 26)^2 = 0 \] \[ 10x^2 + 7x - 26 = 0 \] Now we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \(a = 10\), \(b = 7\), and \(c = -26\).

   First, calculate the discriminant: \[ b^2 - 4ac = 7^2 - 4 \cdot 10 \cdot (-26) = 49 + 1040 = 1089 \] Now plug in the values into the formula: \[ x = \frac{-7 \pm \sqrt{1089}}{2 \cdot 10} \] Calculate \(\sqrt{1089}\): \[ \sqrt{1089} = 33 \] Now we have: \[ x = \frac{-7 \pm 33}{20} \] This gives us two solutions: \[ x_1 = \frac{-7 + 33}{20} = \frac{26}{20} = 1.3 \] \[ x_2 = \frac{-7 - 33}{20} = \frac{-40}{20} = -2 \]

In summary, the solutions are:

• From the first equation: \(x = 1.3\)
• From the second equation: \(x = 1.3\) and \(x = -2\)

So, the final solutions to the equation are: \[ x = 1.3 \quad \text{and} \quad x = -2 \]