How do you write $3 x (10 - 4 x) - 6 (x = 5) + 15$ $3x(10-4x)-6(x=5)+15$ in standard form?

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Answer 1 · 2017-10-24T01:42:44

$- 4 x^{2} + 8 x - 15 = 0$ $-4x^2 + 8x - 15 = 0$

$a x^{2} + b x + c = 9$ $ax^2 + bx + c = 9$ is the standard form of quadratic equation.

Removing braces,

$30 x - 12 x^{2} - 6 x = 30 + 15$ $30x -12x^2 -6x = 30 + 15$

$- 12 x^{2} + 24 x = 45$ $-12x^2 + 24x = 45$

$- 12 x^{2} + 24 x - 45 = 0$ $-12x^2 + 24x -45 = 0$

# dividing both sides by 3,

$- 4 x 2 + 8 x - 15 = 0$ $-4x2 + 8x -15 = 0$

How do you write 3x(10-4x)-6(x=5)+153x(10−4x)−6(x=5)+153x(10-4x)-6(x=5)+15 in standard form?