What are the solutions to the equation quadraticformula x^2-4x=5 ?

The solutions to the equation quadraticformula x^2-4x=5 are x=5,x=-1

Find the zeros of quadraticformula x^2-4x=5

The zeros of quadraticformula x^2-4x=5 are x=5,x=-1

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x^{2} - 4 x = 5

x = 5, x = - 1

Solution steps

x^{2} - 4 x = 5

Move

5

to the left side

x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

x = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

x = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

x^{2} - 2 x = 3

x, 3^{x} = 8

\frac{180}{50}

x, 4 x = 9

144 x^{2}

What are the solutions to the equation quadraticformula x^2-4x=5 ?
The solutions to the equation quadraticformula x^2-4x=5 are x=5,x=-1
Find the zeros of quadraticformula x^2-4x=5
The zeros of quadraticformula x^2-4x=5 are x=5,x=-1