What are the solutions to the equation 7x^2+5-x=6x^2+10 ?

The solutions to the equation 7x^2+5-x=6x^2+10 are x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2

Find the zeros of 7x^2+5-x=6x^2+10

The zeros of 7x^2+5-x=6x^2+10 are x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2

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7 x^{2} + 5 - x = 6 x^{2} + 10

x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

Decimal Notation

x = 2.79128 \dots, x = - 1.79128 \dots

Solution steps

7 x^{2} + 5 - x = 6 x^{2} + 10

Move

10

to the left side

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

Move

6 x^{2}

to the left side

x^{2} - x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 5) = 21

x_{1, 2} = \frac{- ( - 1 ) \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 21}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 21}{2 \cdot 1}

x = \frac{- ( - 1 ) + 21}{2 \cdot 1} : \frac{1 + 21}{2}

x = \frac{- ( - 1 ) - 21}{2 \cdot 1} : \frac{1 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

e^{x} > 7

u^{4} - 11 u^{3} + 10 u^{2}

i^{63}

- 12 x^{2} + 192

x^{2} + 8 x + 13 = 0

What are the solutions to the equation 7x^2+5-x=6x^2+10 ?
The solutions to the equation 7x^2+5-x=6x^2+10 are x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2
Find the zeros of 7x^2+5-x=6x^2+10
The zeros of 7x^2+5-x=6x^2+10 are x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2