What are the solutions to the equation u^2-3u+112=12u+56 ?

The solutions to the equation u^2-3u+112=12u+56 are u=8,u=7

Find the zeros of u^2-3u+112=12u+56

The zeros of u^2-3u+112=12u+56 are u=8,u=7

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u^{2} - 3 u + 112 = 12 u + 56

u = 8, u = 7

Solution steps

u^{2} - 3 u + 112 = 12 u + 56

Move

56

to the left side

u^{2} - 3 u + 56 = 12 u

Move

12 u

to the left side

u^{2} - 15 u + 56 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 56 = 1

u_{1, 2} = \frac{- ( - 15 ) \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 15 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 15 ) - 1}{2 \cdot 1}

u = \frac{- ( - 15 ) + 1}{2 \cdot 1} : 8

u = \frac{- ( - 15 ) - 1}{2 \cdot 1} : 7

The solutions to the quadratic equation are:

u = 8, u = 7

3 x^{2} - 8 x - 20 = - 9

x + 1 = x (x - 4)

x, x^{2} - 50 y^{2} - 10 x + 25 = 0

x^{2} + 24 x + 132

x^{2} + 23 x + 132 = 0

What are the solutions to the equation u^2-3u+112=12u+56 ?
The solutions to the equation u^2-3u+112=12u+56 are u=8,u=7
Find the zeros of u^2-3u+112=12u+56
The zeros of u^2-3u+112=12u+56 are u=8,u=7