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

The solutions to the equation x^2-3x=4 are x=4,x=-1

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

The zeros of x^2-3x=4 are x=4,x=-1

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x^{2} - 3 x = 4

x = 4, x = - 1

Solution steps

x^{2} - 3 x = 4

Move

4

to the left side

x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

(- 2 x - 7) \times 2 x + 9 (- 2 x - 7)

5 x^{2} - 17 x = - 6

6 (x^{2} + 1) - (2 x - 4) (3 x + 2) < 3 (5 x + 21)

\frac{3 ( x - 4 )}{2} + \frac{5 ( 2 x + 1 )}{3} = \frac{2 x}{4}

x^{3} y^{3} + 512

What are the solutions to the equation x^2-3x=4 ?
The solutions to the equation x^2-3x=4 are x=4,x=-1
Find the zeros of x^2-3x=4
The zeros of x^2-3x=4 are x=4,x=-1