Find a
\( 2(a-4)^2+3=163-16a \)
\( -8 \)
\( \pm8 \)
\( 4(a-7)^2=(2a-3)^2 \)
\( 4\frac{1}{4} \)
\( -4\frac{1}{4} \)
\( 7+(x-5)^2=(x+3)(x+3) \)
\( x=-0.04 \)
\( x=1\frac{16}{23} \)
Find a a given that
\( 2a(a-5)=(a+3)^2+(a-3)^2 \)
\( -1.8 \)
\( 0 \)
\( (5-3a)^2+a=(a+1)^2-31a \)
No solution
\( \sqrt{3} \)
2(a−4)2+3=163−16a 2(a-4)^2+3=163-16a 2(a−4)2+3=163−16a
±8 \pm8 ±8
4(a−7)2=(2a−3)2 4(a-7)^2=(2a-3)^2 4(a−7)2=(2a−3)2
414 4\frac{1}{4} 441
7+(x−5)2=(x+3)(x+3) 7+(x-5)^2=(x+3)(x+3) 7+(x−5)2=(x+3)(x+3)
x=11623 x=1\frac{16}{23} x=12316
2a(a−5)=(a+3)2+(a−3)2 2a(a-5)=(a+3)^2+(a-3)^2 2a(a−5)=(a+3)2+(a−3)2
−1.8 -1.8 −1.8
(5−3a)2+a=(a+1)2−31a (5-3a)^2+a=(a+1)^2-31a (5−3a)2+a=(a+1)2−31a