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

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

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

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

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

x = 1, x = - 3

Solution steps

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 2 + 4}{2 \cdot 1} : 1

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

The solutions to the quadratic equation are:

x = 1, x = - 3

w^{2} - 6 w

0 = 3 x^{2} - 8 x

4 x^{2} - 25

y^{2} - 8 y = - 11

2 x^{2} - 16 = - 12 x

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