What are the solutions to the equation x^2=4x+7 ?

The solutions to the equation x^2=4x+7 are x=2+sqrt(11),x=2-sqrt(11)

Find the zeros of x^2=4x+7

The zeros of x^2=4x+7 are x=2+sqrt(11),x=2-sqrt(11)

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x^{2} = 4 x + 7

x = 2 + 11, x = 2 - 11

Decimal Notation

x = 5.31662 \dots, x = - 1.31662 \dots

Solution steps

x^{2} = 4 x + 7

Move

7

to the left side

x^{2} - 7 = 4 x

Move

4 x

to the left side

x^{2} - 7 - 4 x = 0

Write in the standard form

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

x^{2} - 4 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 7) = 211

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 11}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 11}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2 + 11, x = 2 - 11

192 π = 2 π \cdot x \cdot 16 + 2 π \cdot x^{2}

3 x^{2} - 2 x = 40

2 x^{2} + 7 x + 6 = 0

5 x^{2} - 15 x + 1

(0.5 - x) (1.5 \cdot 1 0^{- 5}) = x^{2}

What are the solutions to the equation x^2=4x+7 ?
The solutions to the equation x^2=4x+7 are x=2+sqrt(11),x=2-sqrt(11)
Find the zeros of x^2=4x+7
The zeros of x^2=4x+7 are x=2+sqrt(11),x=2-sqrt(11)