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4x^2+21x=197
We move all terms to the left:
4x^2+21x-(197)=0
a = 4; b = 21; c = -197;
Δ = b2-4ac
Δ = 212-4·4·(-197)
Δ = 3593
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{3593}}{2*4}=\frac{-21-\sqrt{3593}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{3593}}{2*4}=\frac{-21+\sqrt{3593}}{8} $
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