How many real roots are in y=2x^2-7x-15?

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Answer 1 · 2018-03-29T06:14:25

The real roots in $y=2x^2-7x-15$ are $x=-3/2$ and $x=5$ .

$y=2x^2-7x-15$

$y=0$

$2x^2-7x-15=0$

Attempting to factorise

$2x^2+3x-10x-15=0$

$x(2x+3)-5(2x+3)=0$

$(x-5)(2x+3)=0$

$x-5=0$

$x=5$

$2x+3=0$

$2x=-3$

$x=-3/2$

The real roots in $y=2x^2-7x-15$ are $x=-3/2$ and $x=5$ .