Question

Resolva : me ajuda. Acho que tem que ser por baskharas!! Vale 2 pontos.

Answer 1

$4x^2-28x+13=0 \\ \\ \Delta=b^2-4ac \\ \Delta=(-28)^2-4*4*13 \\ \Delta=784-208 \\ \boxed{\Delta=576}$

$x= \frac{ -b+- \sqrt{\Delta} }{2a} \\ \\ x= \frac{ -(-28)+- \sqrt{576} }{2*4} \\ \\ x= \frac{ 28+-24 }{8} \\ \\ \boxed{x'= \frac{28+24}{8} = \frac{52}{8} = \frac{26}{4} = \boxed{\frac{13}{2} }}$

$\boxed{x''= \frac{28-24}{8} = \frac{4}{8} = \boxed{\frac{1}{2} }} \\ \\ \\ \\ \boxed{\boxed{S.: \left\{ \frac{1}{2} ,\ \ \ \frac{13}{2}\right\}}}}$