What are the solutions to the equation 32=x^2-3x+4 ?

The solutions to the equation 32=x^2-3x+4 are x=7,x=-4

Find the zeros of 32=x^2-3x+4

The zeros of 32=x^2-3x+4 are x=7,x=-4

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32 = x^{2} - 3 x + 4

x = 7, x = - 4

Solution steps

32 = x^{2} - 3 x + 4

Switch sides

x^{2} - 3 x + 4 = 32

Move

32

to the left side

x^{2} - 3 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 7, x = - 4

9 x^{2} - 2 = 3 x

4 x^{2} + 8 x = 7

x, x^{2} y - 9 y - 3 x^{2} = 0

4 x^{2} - x = 0

A, A + BC = (A + B) (A + C)

What are the solutions to the equation 32=x^2-3x+4 ?
The solutions to the equation 32=x^2-3x+4 are x=7,x=-4
Find the zeros of 32=x^2-3x+4
The zeros of 32=x^2-3x+4 are x=7,x=-4