Solve 10-0*1^2+12-0*2I+13-0*3I-5I=0

$$10-0 \cdot { 1 }^{ 2 } +12-0 \cdot 2I+13-0 \cdot 3I-5I = 0$$

Solve for I

$$I=7$$

Steps for Solving Linear Equation

Calculate 1 to the power of 2 and get 1.

$$10-0\times 1+12-0\times 2I+13-0\times 3I-5I=0$$

Multiply 0 and 1 to get 0.

$$10-0+12-0\times 2I+13-0\times 3I-5I=0$$

Subtract 0 from 10 to get 10.

$$10+12-0\times 2I+13-0\times 3I-5I=0$$

Add 10 and 12 to get 22.

$$22-0\times 2I+13-0\times 3I-5I=0$$

Multiply 0 and 2 to get 0.

$$22-0I+13-0\times 3I-5I=0$$

Anything times zero gives zero.

$$22-0+13-0\times 3I-5I=0$$

Subtract 0 from 22 to get 22.

$$22+13-0\times 3I-5I=0$$

Add 22 and 13 to get 35.

$$35-0\times 3I-5I=0$$

Multiply 0 and 3 to get 0.

$$35-0I-5I=0$$

Anything times zero gives zero.

$$35-0-5I=0$$

Subtract 0 from 35 to get 35.

$$35-5I=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$-5I=-35$$

Divide both sides by -5.

$$I=\frac{-35}{-5}$$

Divide -35 by -5 to get 7.

$$I=7$$