What are the solutions to the equation quadraticformula x^2=11x-28 ?

The solutions to the equation quadraticformula x^2=11x-28 are x=7,x=4

Find the zeros of quadraticformula x^2=11x-28

The zeros of quadraticformula x^2=11x-28 are x=7,x=4

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x^{2} = 11 x - 28

x = 7, x = 4

Solution steps

x^{2} = 11 x - 28

Move

28

to the left side

x^{2} + 28 = 11 x

Move

11 x

to the left side

x^{2} + 28 - 11 x = 0

Write in the standard form

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

x^{2} - 11 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

x_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

x = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

x = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 7, x = 4

∣ - 15 - 3 ∣

x^{2} = 5 x - 6

- 63/7

36 x^{2}

\frac{8}{( 2 )}

What are the solutions to the equation quadraticformula x^2=11x-28 ?
The solutions to the equation quadraticformula x^2=11x-28 are x=7,x=4
Find the zeros of quadraticformula x^2=11x-28
The zeros of quadraticformula x^2=11x-28 are x=7,x=4