Factorise the given statement :Given, five times the number = 3 less than twice the square of the number.

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Solution

$5 x = 2 x^{2} - 3$
$2 x^{2} - 5 x - 3$
$2 x^{2} - 5 x - 3 = 0$
$2 x^{2} + x - 6 x - 3 = 0$
$x (2 x + 1) - 3 (2 x + 1) = 0$
$(x - 3) (2 x + 1) = 0$
$x = 3, x = - \frac{1}{2}$
$∴ number = 3$

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Factorise the given statement :Given, five times the number = 3 less than twice the square of the number.

5x=2x2−32x2−5x−32x2−5x−3=02x2+x−6x−3=0x(2x+1)−3(2x+1)=0(x−3)(2x+1)=0x=3,x=−12∴number = 3

$5 x = 2 x^{2} - 3$
$2 x^{2} - 5 x - 3$
$2 x^{2} - 5 x - 3 = 0$
$2 x^{2} + x - 6 x - 3 = 0$
$x (2 x + 1) - 3 (2 x + 1) = 0$
$(x - 3) (2 x + 1) = 0$
$x = 3, x = - \frac{1}{2}$
$∴ number = 3$