What are the solutions to the equation y^2-4y=21 ?

The solutions to the equation y^2-4y=21 are y=7,y=-3

Find the zeros of y^2-4y=21

The zeros of y^2-4y=21 are y=7,y=-3

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y^{2} - 4 y = 21

y = 7, y = - 3

Solution steps

y^{2} - 4 y = 21

Move

21

to the left side

y^{2} - 4 y - 21 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

y_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

y = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

y = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

y = 7, y = - 3

\frac{2 m n ^{2} ( 3 m ^{2} n ) ^{2}}{12 m ^{3} n ^{4}}

lo g_{3} (\frac{1}{27}) = x

- \frac{2}{3} - \frac{7}{9}

∣ 3.1416 - 7 ∣ > 3

(2 i - 3) - (7 + \frac{1}{5} i)

What are the solutions to the equation y^2-4y=21 ?
The solutions to the equation y^2-4y=21 are y=7,y=-3
Find the zeros of y^2-4y=21
The zeros of y^2-4y=21 are y=7,y=-3