If a quadratic equation has solutions [tex] x_1 [/tex] and [tex] x_2 [/tex], then we can write it as
[tex] (x-x_1)(x-x_2) [/tex]
So, in your case, we have (let me write 1.5 as 3/2 and -0.25 as -1/4)
[tex] \left(x-\dfrac{3}{2}\right)\left(x+\dfrac{1}{4}\right) = x^2-\dfrac{5}{4}x-\dfrac{3}{8}[/tex]
If we don't want decimal coefficient, we can multiply the whole expression by 8:
[tex]8x^2-10x-3[/tex]
