What are the solutions to the equation 12^2=(10+x)(x) ?

The solutions to the equation 12^2=(10+x)(x) are x=8,x=-18

Find the zeros of 12^2=(10+x)(x)

The zeros of 12^2=(10+x)(x) are x=8,x=-18

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1 2^{2} = (10 + x) (x)

x = 8, x = - 18

Solution steps

1 2^{2} = (10 + x) (x)

Expand

1 2^{2} : 144

Expand

(10 + x) (x) : 10 x + x^{2}

144 = 10 x + x^{2}

Switch sides

10 x + x^{2} = 144

Move

144

to the left side

10 x + x^{2} - 144 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + 10 x - 144 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 144 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 144) = 26

x_{1, 2} = \frac{- 10 \pm 26}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 26}{2 \cdot 1}, x_{2} = \frac{- 10 - 26}{2 \cdot 1}

x = \frac{- 10 + 26}{2 \cdot 1} : 8

x = \frac{- 10 - 26}{2 \cdot 1} : - 18

The solutions to the quadratic equation are:

x = 8, x = - 18

b, a^{3} - a b^{2} - a c^{2} - ab c + b^{2} c + b c^{2} = 0

x^{2} - 13 + 12 = 0

8 x^{2} - 10 x + 2 = 0

x, (x - 3)^{2} + (y + 5)^{2} + z^{2} = 81

3 (2 x + 5)^{2} = (2 x - 5)^{2} + 10

What are the solutions to the equation 12^2=(10+x)(x) ?
The solutions to the equation 12^2=(10+x)(x) are x=8,x=-18
Find the zeros of 12^2=(10+x)(x)
The zeros of 12^2=(10+x)(x) are x=8,x=-18