What are the solutions to the equation y^2=13y-42 ?

The solutions to the equation y^2=13y-42 are y=7,y=6

Find the zeros of y^2=13y-42

The zeros of y^2=13y-42 are y=7,y=6

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y^{2} = 13 y - 42

y = 7, y = 6

Solution steps

y^{2} = 13 y - 42

Move

42

to the left side

y^{2} + 42 = 13 y

Move

13 y

to the left side

y^{2} + 42 - 13 y = 0

Write in the standard form

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

y^{2} - 13 y + 42 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 42 = 1

y_{1, 2} = \frac{- ( - 13 ) \pm 1}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 13 ) + 1}{2 \cdot 1}, y_{2} = \frac{- ( - 13 ) - 1}{2 \cdot 1}

y = \frac{- ( - 13 ) + 1}{2 \cdot 1} : 7

y = \frac{- ( - 13 ) - 1}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

y = 7, y = 6

x^{2} + 20 x - 96 = 0

9 x^{2} + 12 x + 5 = 10

v, p^{1} v^{1} = p^{2} v^{2}

3 x^{2} - 5 x = 0

x, x^{2} - 3 m x + 8 = 0

What are the solutions to the equation y^2=13y-42 ?
The solutions to the equation y^2=13y-42 are y=7,y=6
Find the zeros of y^2=13y-42
The zeros of y^2=13y-42 are y=7,y=6