What are the solutions to the equation (4x-7)^2+16=40x-70 ?

The solutions to the equation (4x-7)^2+16=40x-70 are x= 15/4 ,x= 9/4

Find the zeros of (4x-7)^2+16=40x-70

The zeros of (4x-7)^2+16=40x-70 are x= 15/4 ,x= 9/4

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(4 x - 7)^{2} + 16 = 40 x - 70

x = \frac{15}{4}, x = \frac{9}{4}

Decimal Notation

x = 3.75, x = 2.25

Solution steps

(4 x - 7)^{2} + 16 = 40 x - 70

Expand

(4 x - 7)^{2} + 16 : 16 x^{2} - 56 x + 65

16 x^{2} - 56 x + 65 = 40 x - 70

Move

70

to the left side

16 x^{2} - 56 x + 135 = 40 x

Move

40 x

to the left side

16 x^{2} - 96 x + 135 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 96 ) \pm ( - 96 ) ^{2} - 4 \cdot 16 \cdot 135}{2 \cdot 16}

(- 96)^{2} - 4 \cdot 16 \cdot 135 = 24

x_{1, 2} = \frac{- ( - 96 ) \pm 24}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 96 ) + 24}{2 \cdot 16}, x_{2} = \frac{- ( - 96 ) - 24}{2 \cdot 16}

x = \frac{- ( - 96 ) + 24}{2 \cdot 16} : \frac{15}{4}

x = \frac{- ( - 96 ) - 24}{2 \cdot 16} : \frac{9}{4}

The solutions to the quadratic equation are:

x = \frac{15}{4}, x = \frac{9}{4}

\frac{12}{9}

x^{2} - 9 < 0.1

\frac{200 x ^{3} y}{2 x y ^{5}}

16 s^{4} + 36 s^{3} b - 10 s^{2} b^{2}

y^{- 4}

What are the solutions to the equation (4x-7)^2+16=40x-70 ?
The solutions to the equation (4x-7)^2+16=40x-70 are x= 15/4 ,x= 9/4
Find the zeros of (4x-7)^2+16=40x-70
The zeros of (4x-7)^2+16=40x-70 are x= 15/4 ,x= 9/4