1
Move everything to the left by subtracting 25 from both sides:
[tex]81x^2-25=0[/tex]
Both terms are squares: [tex]81x^2[/tex] is the square of [tex]9x[/tex], while 25 is the square of 5. So, we can apply the factoring
[tex]a^2-b^2=(a+b)(a-b)[/tex]
To factor the expression as
[tex](9x+5)(9x-5)=0[/tex]
A multiplication is zero if and only if one of the factors is zero, so we have
[tex](9x+5)(9x-5)=0 \iff 9x+5=0 \lor 9x-5=0 \iff x=-\dfrac{5}{9} \lor x=\dfrac{5}{9}[/tex]
2
The discriminant of the quadratic expression [tex]ax^2+bx+c[/tex] is defined as
[tex]\Delta = b^2-4ac[/tex]
So, you have
[tex]\Delta = 0^2-4\cdot(-1)\cdot(-25) = -100[/tex]
3
Similarly, we have
[tex]\Delta = 7^2-4\cdot(-3)\cdot 6 = 49+72=121[/tex]
