Respuesta :
Before doing the question, let's recall the quadratic formula which is used to find the roots of a quadratic equation ;
For any quadratic equation of the form ax² + bx + c = 0 , it's roots are given by the quadratic formula as :
- [tex]{\boxed{\bf{x=\dfrac{-b\pm \sqrt{D}}{2a}}}}[/tex]
Where , D = b² - 4ac (Discriminant)
Now , simplifying the equation :
[tex]{:\implies \quad \sf -3\{(x)^{2}+2^{2}-2\times 2\times x\}+17=0\quad \qquad \{\because (a-b)^{2}=a^{2}+b^{2}-2ab\}}[/tex]
[tex]{:\implies \quad \sf -3(x^{2}+4-4x)+17=0}[/tex]
[tex]{:\implies \quad \sf -3x^{2}-12+12x+17=0}[/tex]
[tex]{:\implies \quad \sf -3x^{2}+12x+5=0}[/tex]
Dividing both sides by -1 will yield :
[tex]{:\implies \quad \sf 3x^{2}-12x-5=0}[/tex]
Now , it's in the form of the standard quadratic equation , where a = 3 , b = -12 , c = 5 , and D = (-12)² - 4 × 3 × -5 = 144+60 = 204
Now, By quadratic Formula ;
[tex]{:\implies \quad \sf x=\dfrac{-(-12)\pm \sqrt{204}}{2\times 3}}[/tex]
[tex]{:\implies \quad \sf x=\dfrac{12\pm 2\sqrt{51}}{2\times 3}}[/tex]
[tex]{:\implies \quad \sf x=\dfrac{\cancel{2}(6\pm \sqrt{51})}{\cancel{2}\times 3}}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{x=\dfrac{6\pm \sqrt{51}}{3}}}}[/tex]
