What are the solutions to the equation (u-14)(u+10)=-104 ?

The solutions to the equation (u-14)(u+10)=-104 are u=2(1+sqrt(10)),u=2(1-sqrt(10))

Find the zeros of (u-14)(u+10)=-104

The zeros of (u-14)(u+10)=-104 are u=2(1+sqrt(10)),u=2(1-sqrt(10))

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(u - 14) (u + 10) = - 104

u = 2 (1 + 10), u = 2 (1 - 10)

Decimal Notation

u = 8.32455 \dots, u = - 4.32455 \dots

Solution steps

(u - 14) (u + 10) = - 104

Expand

(u - 14) (u + 10) : u^{2} - 4 u - 140

u^{2} - 4 u - 140 = - 104

Move

104

to the left side

u^{2} - 4 u - 36 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 36) = 410

u_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 4 10}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 4 10}{2 \cdot 1}

u = \frac{- ( - 4 ) + 4 10}{2 \cdot 1} : 2 (1 + 10)

u = \frac{- ( - 4 ) - 4 10}{2 \cdot 1} : 2 (1 - 10)

The solutions to the quadratic equation are:

u = 2 (1 + 10), u = 2 (1 - 10)

\frac{1}{2} - \frac{3}{2} i

\frac{b}{30} = \frac{4}{5}

x + 5.5 + 8 = 5 x - 13.5 - 4 x

a x + b x - a y - b y

(2 x - 13)^{2} = 8

What are the solutions to the equation (u-14)(u+10)=-104 ?
The solutions to the equation (u-14)(u+10)=-104 are u=2(1+sqrt(10)),u=2(1-sqrt(10))
Find the zeros of (u-14)(u+10)=-104
The zeros of (u-14)(u+10)=-104 are u=2(1+sqrt(10)),u=2(1-sqrt(10))